Solve (sqrt{2}-1)^2+sqrt{3}*(sqrt{3}-sqrt{6})+sqrt{8}

$$( \sqrt { 2 } - 1 ) ^ { 2 } + \sqrt { 3 } \times ( \sqrt { 3 } - \sqrt { 6 } ) + \sqrt { 8 }$$

Evaluate

$$6-3\sqrt{2}\approx 1.757359313$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}-1\right)^{2}.

$$\left(\sqrt{2}\right)^{2}-2\sqrt{2}+1+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

The square of \sqrt{2} is 2.

$$2-2\sqrt{2}+1+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

Add 2 and 1 to get 3.

$$3-2\sqrt{2}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}$$

Use the distributive property to multiply \sqrt{3} by \sqrt{3}-\sqrt{6}.

$$3-2\sqrt{2}+\left(\sqrt{3}\right)^{2}-\sqrt{3}\sqrt{6}+\sqrt{8}$$

The square of \sqrt{3} is 3.

$$3-2\sqrt{2}+3-\sqrt{3}\sqrt{6}+\sqrt{8}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$3-2\sqrt{2}+3-\sqrt{3}\sqrt{3}\sqrt{2}+\sqrt{8}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$3-2\sqrt{2}+3-3\sqrt{2}+\sqrt{8}$$

Add 3 and 3 to get 6.

$$6-2\sqrt{2}-3\sqrt{2}+\sqrt{8}$$

Combine -2\sqrt{2} and -3\sqrt{2} to get -5\sqrt{2}.

$$6-5\sqrt{2}+\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$6-5\sqrt{2}+2\sqrt{2}$$

Combine -5\sqrt{2} and 2\sqrt{2} to get -3\sqrt{2}.

$$6-3\sqrt{2}$$